Question: A person stands $15$ meters east of an intersection and watches a car driving towards the intersection from the north at $1$ meter per second. At a certain instant, the car is $8$ meters from the intersection. What is the rate of change of the distance between the car and the person at that instant (in meters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{8}{17}$ (Choice B) B $-17$ (Choice C) C $-2.125$ (Choice D) D $-\sqrt{65}$
Solution: Setting up the math Let... $a(t)$ denote the distance between the car and the intersection at time $t$, $b$ denote the distance between the person and the intersection (which is always $15$ meters), and $c(t)$ denote the distance between the car and the person at time $t$. $a(t)$ $b$ $c(t)$ We are given that $a'(t)=-1$ and $b=15$. Notice that $a'$ is negative since the car is getting closer to the intersection. We are also given that $a(t_0)=8$ for a specific time $t_0$. We want to find $c'(t_0)$. Relating the measures The measures relate to each other through the Pythagorean theorem: $\begin{aligned} \,[a(t)]^2+b^2&=[c(t)]^2 \\\\\\ [a(t)]^2+(15)^2&=[c(t)]^2 \end{aligned}$ We can differentiate both sides to find an expression for $c'(t)$ : $c'(t)=\dfrac{a(t)a'(t)}{c(t)}$ Using the information to solve In order to find $c'(t_0)$ we need to find $c(t_0)$. Using the Pythagorean theorem and the fact that $a(t_0)=8$ and $b=15$, we can find that $c(t_0)=17$. Now let's plug ${a(t_0)}={8}$, ${a'(t_0)}={-1}$, and ${c(t_0)}={17}$ into the expression for $c'(t_0)$ : $\begin{aligned} c'(t_0)&=\dfrac{{a(t_0)}{a'(t_0)}}{{c(t_0)}} \\\\ &=\dfrac{({8})({-1})}{({17})} \\\\ &=-\dfrac{8}{17} \end{aligned}$ In conclusion, the rate of change of the distance between the car and the person at that instant is $-\dfrac{8}{17}$ meters per second. Since the rate of change is negative, we know that the distance is decreasing.